Here's a proof since I think the current answers are possibly a little lacking (and the approach taken in the question itself seems slightly off):
As in the question, let $A = k[X_0,\dots,X_n]$, $X = V(I)$ and $H = V(f)$. For a closed subset $Z = V(J) \subset \mathbb{P}^n_k$, write $\overline{Z}=V(J)\subset \mathbb{A}^{n+1}_k$. First we proceed to show that $\overline{X}\cap\overline{H}(=\overline{X\cap H})$ has dimension at least $1$.
Notice that $\overline{X\cap H} = V(I+(f))$, and since $I$ and $f$ are homogeneous, we have $I+(f) \subset (X_0,\dots,X_n)$, so there is some prime $I+(f) \subset \mathfrak{p} \subset (X_0,\dots,X_n)$ minimal over $I+(f)$ (corresponding to an irreducible component of the affine cone over the intersection of $X$ and $H$). If $\mathfrak{p} \subsetneq (X_0,\dots,X_n)$, then $V(I+(f))$ has dimension at least $1$, as claimed. If not, $\mathfrak{p} = (X_0,\dots,X_n)$ and so by Krull's principal ideal theorem applied to $A/I$, any chain of primes between $I$ and $(X_0,\dots,X_n)$ has length at most $1$. But since $X$ has dimension at least $1$, there are homogeneous primes $I \subset \mathfrak{q}\subsetneq\mathfrak{q}'$ not containing $(X_0,\dots,X_n)$, giving a chain of primes from $I$ to $(X_0,\dots,X_n)$ of length $2$, a contradiction.
This dimension argument gives that there is a prime ideal inbetween $I+(f)$ and $(X_0,\dots X_n)$, and so the question is: is there a homogeneous prime ideal between them? There's a high tech way and a low tech way to proceed, but know that the high tech approach is really just hiding the low tech one behind some fancy language.
High tech: We have a map $\pi:\mathbb{A}^{n+1}_k\setminus\{0\}\rightarrow\mathbb{P}^n_k$, such that for any closed subset $Z$, $\pi^{-1}(Z) = \overline{Z}\setminus\{0\}$, so the cone over the intersection having dimension $1$ means that it contains a non-zero point, and so this maps to the intersection under $\pi$, meaning that the intersection is therefore non-empty.
Low tech: In the process of the $\rm{Proj}(S_{\bullet})$ construction, we get a bijection between the primes of $S_{(f)}$ and the homogeneous primes of $S_{f}$, where the map one way is given by taking the degree $0$ component. But in fact, given any prime ideal of $S_f$, taking the degree $0$ component gives a prime ideal of $S_{(f)}$ and so, by the bijection, there is some homogenous prime of $S_f$ with the same degree $0$ component. Thus, in our construction, the prime $\mathfrak{p}\subsetneq (X_0,\dots, X_n)$ lying over $I+(f)$ doesn't contain some $X_i$, and so corresponds to a prime in $A_{X_i}$ lying over $(I+(f))_{X_i}$. Then by the remarks above there is then a homogeneous prime of $A_{X_i}$ with the same degree $0$ component, which therefore also contains $(I+(f))_{X_i}$. This prime then corresponds to a homogeneous prime of $A$ (also not containing $X_i$) that contains $I+(f)$, as desired. Actually checking that all of the correspondences work out as I claim is somewhat of a hassle (it's not hard per say, but is a bit of a faff) but bear in mind that you would have to do all of this anyway if you wanted to prove that the projection map used in the high tech approach actually has the desired property, so I don't think it can really be avoided.
Let $Y_0\subsetneq Y_1 \subsetneq\cdots\subsetneq Y_n$ be an arbitrary chain of irreducible, closed subsets of $Y$. Let $\overline{Y_i}$ denote the closure of $Y_i$ in $X$. $\overline{ Y_{i-1}}\subsetneq \overline{ Y_i}$, since, otherwise, $Y_{i-1}=Y\cap \overline{Y_{i-1}}=Y\cap \overline{Y_i}=Y_i,$ contrary to the hypothesis.
If $\overline{Y_i}$ is irreducible in $X$ by Proposition 1.6 of Robin Hartshorne's "Algebraic Geometry":
Any nonempty open subset of an irreducible topological space is dense and irreducible, If $Y$ is a subset of a topological space $X$, which is irreducible in its induced topology, then the closure $\overline Y$ is also irreducible.
Hence, we have created a chain $\overline{Y_0}\subsetneq \overline{Y_1} \subsetneq\cdots\subsetneq \overline{Y_n}$ in $X$. By the definition of $X$'s dimension, $n\le \dim X$. Since $Y_0\subsetneq Y_1 \subsetneq\cdots\subsetneq Y_n$ was arbitrary, $\dim Y\le \dim X.$
Best Answer
Your solution is correct (assuming you can also prove that $\pi(Z)$ really is a hypersurface). I'm not sure what's going on with the hint--Exercise 11.1.E applies just as well to directly show $\dim Z=\dim \pi(Z)$, and indeed the argument you have made is essentially just reproving Exercise 11.1.E in that particular case.